Question

A simple pendulum with iron bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of liquid is $\frac{1}{12}$ th that of iron, then new time period will be

• A. $T \sqrt{\frac{8}{7}}$
• B. $T \sqrt{\frac{12}{13}}$
• C. $T \sqrt{\frac{12}{11}}$
• D. $T \sqrt{\frac{6}{5}}$

Answer 1

Answer: (C)

When bob is inserted, in liquid effective of is reduced because of the force of upthrust.
$\rho_{s}$ is density of solid.
$\rho_{L}$ is density of liquid.
$V$ is volume of solid
and $T=2 \pi \sqrt{\frac{l}{g}}$
$T_{\text {new }} =2 \pi \sqrt{\frac{l}{g_{\text {new }}}}$
$ T_{\text {new }} =2 \pi \sqrt{\frac{l}{g-\frac{\rho_{L}}{\rho_{S}} g}} $
$ T_{\text {new }} =2 \pi \sqrt{\frac{l \times 12}{g \times 11}} $
or $ T_{\text {new }} =\sqrt{\frac{12}{11}} T$